Number of roots of tanh(ax) = x

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In summary, when the value of beta is varied from 1/2 to 3/2, the number of solutions of the equation x = tanh(beta x) can either remain unchanged or increase by 1, 2, or 3 depending on the value of beta. The shape of tanh(beta x) is nearly a straight line near the origin with horizontal asymptotes at y = 1 and y = -1. The slope of the line tangent to tanh(beta x) at the origin is beta, and for beta < 1, the slope keeps decreasing and never intersects y = x again. However, for beta > 1, tanh(beta x) intersects y = x again at
  • #1
shakgoku
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Homework Statement



what happens to the number of solutions of the equation
[tex]x = \tanh(\beta x) [/tex]

When
[tex] \beta[/tex] is varied from [tex]\frac{1}{2} [/tex] to [tex] \frac{3}{2} [/tex]
[/tex]



a) unchanged
b) increase by 1
c) increase by 2
d) increase by 3

Homework Equations



[tex] \tanh(ax) =.... -\frac{17}{315} \, a^{7} x^{7} + \frac{2}{15} \, a^{5} x^{5} -
\frac{1}{3} \, a^{3} x^{3} + a x[/tex]


[tex] tanh(ax)= \frac{e^{ax}-e^{-ax}}{e^{ax}+e^{-ax}}
[/tex]

The Attempt at a Solution



Tried to Apply the two above formulas without any success. This is an examination question and too bad, Graphing and programming calculators are not allowed :(
 
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  • #2
hi shakgoku! :smile:

what's the shape of y = tanhßx?

what's the shape of y = x? :wink:
 
  • #3
hi tiny-tim,

The shape of tanh(βx) is nearly straight line near origin and has horizontal asymptotes at y = 1 and y = -1
The shape of x is straight line through origin , slope = 45 degrees.

so x = 0 is definitely a root. how to find it there are more roots?
 
  • #4
hi shakgoku! :smile:

(just got up :zzz: …)
shakgoku said:
The shape of tanh(βx) is nearly straight line near origin and has horizontal asymptotes at y = 1 and y = -1

yes, but which ways does it bend? :wink:
 
  • #5
What is the slope of the line tangent to tanh(ßx) at the origin?
 
  • #6
What is the slope of the line tangent to tanh(ßx) at the origin?

yes, but which ways does it bend?

Thanks , its so obvious now. The slope is at x = 0 is beta. So, if beta<1 the slope keeps decreasing and never gets a chance to intersect y = x again.

But beta > 1, tanh(beta*x) stays > (y = x) but due to decreasing slope, intersects y = x again at two places x = + and - r
(if r is a root).


I've also found out that 0 < r < 1

by taking f = tanh(beta*x) - x and checking its value for 0 and 1. for one value its +ve and for other it was -ve.
 
  • #7
shakgoku said:
Thanks , its so obvious now. The slope is at x = 0 is beta. So, if beta<1 the slope keeps decreasing and never gets a chance to intersect y = x again.

But beta > 1, tanh(beta*x) stays > (y = x) but due to decreasing slope, intersects y = x again at two places x = + and - r

yup!

just sometimes we can prove something that looks complicated with hardly any maths! :biggrin:
 
  • #8
hi!
I didnt get that beta>1 part. You mean to say, the equation has 3 roots-0,1,-1? Please let me know, am too stuck with the same problem.
 
  • #9
draw the graph!

what does it look like?​
 

FAQ: Number of roots of tanh(ax) = x

What is the equation for the number of roots of tanh(ax) = x?

The equation for the number of roots of tanh(ax) = x is n = 2aπ, where n is the number of roots and a is a constant.

How do you determine the number of roots of tanh(ax) = x?

The number of roots of tanh(ax) = x can be determined by solving the equation n = 2aπ, where n is a positive integer. This means that the number of roots is directly proportional to the value of a.

What is the significance of the number of roots in tanh(ax) = x?

The number of roots in tanh(ax) = x represents the number of times the graph of the function crosses the x-axis. It also indicates the number of solutions to the equation tanh(ax) = x.

Can the number of roots in tanh(ax) = x be negative?

No, the number of roots in tanh(ax) = x cannot be negative as it represents a physical quantity (number of intersections/solutions) and cannot be negative.

How does changing the value of a affect the number of roots in tanh(ax) = x?

The value of a directly affects the number of roots in tanh(ax) = x. As the value of a increases, the number of roots also increases, following the equation n = 2aπ. As a approaches 0, the number of roots decreases and for a = 0, there are no roots.

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